RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Exercise 20C

Exercise 20C

Page-237

Question 1:

Find the area of a parallelogram with base 32 cm and height 16.5 cm.

Answer 1:

Base = 32 cm
Height = 16.5 cm

∴ Area of the parallelogram = Base $\times$ Height
                                          =  32 cm $\times$ 16.5 cm
                                          = 528 cm²

Question 2:

The base of a parallelogram measures 1 m 60 cm and its height is 75 cm. Find its area in m².

Answer 2:

Base = 1 m 60 cm = 1.6 m             [since 100 cm = 1 m]
Height = 75 cm = 0.75 m

∴ Area of the parallelogram = Base $\times$ Height
                                              = 1.6 m $\times$ 0.75 m
                                              = 1.2 m²

Question 3:

In a parallelogram it is being given that base = 14 dm and height = 6.5 dm. Find its area in

(i) cm².
(ii) m².

Answer 3:

(i) Base = 14 dm = (14 $\times$ 10) cm = 140 cm                  [since 1 dm  = 10 cm]
     Height = 6.5 dm = (6.5 $\times$ 10) cm = 65 cm
    
     Area of the parallelogram  = Base $\times$ Height
                                                = 140 cm $\times$ 65 cm
                                                = 9100 cm²

(ii) Base = 14 dm = (14 $\times$ 10) cm                            [since 1 dm = 10 cm and 100 cm  = 1 m]
              = 140 cm = 1.4 m            
      Height = 6.5 dm = (6.5 $\times$ 10) cm
                  = 65 cm = 0.65 m
    
 ∴ Area of the parallelogram = Base $\times$ Height
                                                = 1.4 m $\times$ 0.65 m
                                                = 0.91 m²

Question 4:

Find the height of a parallelogram whose area is 54 cm² and the base is 15 cm.

Answer 4:

Area of the given parallelogram = 54 cm²
Base of the given parallelogram = 15 cm
∴ Height of the given parallelogram = $\frac{\mathrm{Area}}{\mathrm{Base}}$ = $\left(\frac{54}{15}\right)$ cm = 3.6 cm

Question 5:

One side of a parallelogram is 18 cm long and its area is 153 cm². Find the distance of the given side from its opposite side.

Answer 5:

Base of the parallelogram = 18 cm
Area of the parallelogram = 153 cm²
∴ Area of the parallelogram = Base $\times$ Height
⇒ Height = $\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{parallelogram}}{\mathrm{Base}}$ = $\left(\frac{153}{18}\right)$ cm = 8.5 cm
Hence, the distance of the given side from its opposite side is 8.5 cm.

Question 6:

In a parallelogram ABCD, AB = 18 cm, BC = 12 cm. AL ⊥ DC and AM ⊥ BC.


If AL = 6.4 cm, find the length of AM.

Answer 6:

Base, AB = 18 cm
Height, AL = 6.4 cm
∴ Area of the parallelogram ABCD = Base $\times$ Height
                                                    = (18 cm $\times$ 6.4 cm) = 115.2 cm²                    ... (i)

Now, taking BC as the base:
Area of the parallelogram ABCD = Base $\times$ Height
                                                      = (12 cm $\times$ AM)                          ... (ii)
From equation (i) and (ii):
12 cm $\times$ AM = 115.2 cm²
⇒ AM = $\left(\frac{115.2}{12}\right)$cm
= 9.6 cm

Question 7:

The adjacent sides of a parallelogram are 15 cm and 8 cm. If the distance between the longer sides is 4 cm, find the distance between the shorter sides.

Answer 7:

ABCD is a parallelogram with side AB of length 15 cm and the corresponding altitude AE of length 4 cm.
The adjacent side AD is of length 8 cm and the corresponding altitude is CF.

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases.

∴ AD $\times$ CF = AB $\times$ AE
⇒ 8 cm $\times$ CF = 15 cm $\times$4 cm

⇒ CF =$\left(\frac{15\times 4}{8}\right)$ cm = $\left(\frac{15}{2}\right)$ cm = 7.5 cm
Hence, the distance between the shorter sides is 7.5 cm.

Question 8:

The height of a parallelogram is one-third of its base. If the area of the parallelogram is 108 cm², find its base and height.

Answer 8:

Let the base of the parallelogram be x cm.
Then, the height of the parallelogram will be $\frac{1}{3}$x cm.
It is given that the area of the parallelogram is 108 cm².

Area of a parallelogram =  Base $\times$ Height
                     ∴ 108 cm² = x $\times$ $\frac{1}{3}$x
                        108 cm² = $\frac{1}{3}$x²
⇒ x² = (108 $\times$ 3) cm² = 324 cm²
⇒ x² = (18 cm)²
⇒ x = 18 cm

∴ Base = x = 18 cm
Height = $\frac{1}{3}$x = $\left(\frac{1}{3}\times 18\right)$ cm
            = 6 cm

Question 9:

The base of a parallelogram is twice its height. If the area of the parallelogram is 512 cm². find the base and the height.

Answer 9:

Let the height of the parallelogram be x cm.
Then, the base of the parallelogram will be 2x cm.
It is given that the area of the parallelogram is 512 cm².

Area of a parallelogram =  Base $\times$ Height
                     ∴ 512 cm² = 2x $\times$ x
                        512 cm² = 2x²
⇒ x² =$\left(\frac{512}{2}\right)$ cm² = 256 cm²
⇒ x² = (16 cm)²
⇒ x = 16 cm

∴ Base = 2x = 2 $\times$ 16
             = 32 cm
Height = x = 16 cm

Question 10:

Find the area of a rhombus in which

(i) each side = 12 cm and height = 7.5 cm.
(ii) each side = 2 dm and height = 12.6 cm.

Answer 10:

A rhombus is a special type of a parallelogram.

The area of a parallelogram is given by the product of its base and height.
∴ Area of the given rhombus = Base × Height
(i) Area of the rhombus = 12 cm $\times$ 7.5 cm = 90 cm²

(ii) Base = 2 dm = (2 $\times$ 10) = 20 cm    [since 1 dm = 10 cm]
     Height = 12.6 cm
      ∴ Area of the rhombus = 20 cm $\times$ 12.6 cm = 252 cm²

Question 11:

Find the area of a rhombus, the lengths of whose diagonals are:

(i) 16 cm and 28 cm,
(ii) 8 dm 5 cm and 5 dm 6 cm.

Answer 11:

(i)
     Length of one diagonal = 16 cm
     Length of the other diagonal = 28 cm
     ∴ Area of the rhombus = $\frac{1}{2}$ $\times$ (Product of the diagonals)
                                           = $\left(\frac{1}{2}\times 16\times 28\right)$ cm² = 224 cm²
(ii)
      Length of one diagonal = 8 dm 5 cm = (8 $\times$ 10 + 5) cm = 85 cm              [since 1 dm = 10 cm]
      Length of the other diagonal = 5 dm 6 cm = (5 $\times$ 10 + 6) cm = 56 cm
      ∴ Area of the rhombus = $\frac{1}{2}$ $\times$ (Product of the diagonals)
                                            = $\left(\frac{1}{2}\times 85\times 56\right)$ cm²
                                            = 2380 cm²
    

Question 12:

Find the area of a rhombus each side of which measures 20 cm and one of whose diagoanls is 24 cm.

Answer 12:

Let ABCD be the rhombus, whose diagonals intersect at O.

 AB = 20 cm and AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.

Therefore, ΔAOB is a right angled triangle, right angled at O. 

Here, OA =$\frac{1}{2}\mathrm{AC}$ = 12 cm
AB = 20 cm

By Pythagoras theorem:
(AB)² = (OA)² + (OB)²
⇒ (20)² = (12)² + (OB)²
⇒ (OB)² = (20)² − (12)²
⇒ (OB)² = 400 − 144 = 256
⇒ (OB)² = (16)²
⇒ OB = 16 cm
∴ BD = 2 $\times$ OB = 2 $\times$ 16 cm = 32 cm

∴  Area of the rhombus ABCD = $\left(\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}\right)$ cm²
                                                  = $\left(\frac{1}{2}\times 24\times 32\right)$ cm²
                                                  = 384 cm²

Question 13:

The area of a rhombus is 148.8 cm². If one of its diagonals is 19.2 cm, find the length of the other diagonal.

Answer 13:

Area of a rhombus =  $\frac{1}{2}$ $\times$ (Product of the diagonals)
Given:
Length of one diagonal = 19.2 cm
Area of the rhombus = 148.8 cm²

∴ Length of the other diagonal = $\left(\frac{148.8\times 2}{19.2}\right)$ cm = 15.5 cm

Question 14:

The area of a rhombus is 119 cm² and its perimeter is 56 cm. Find its height.

Answer 14:

Perimeter of the rhombus = 56 cm
Area of the rhombus = 119 cm²
Side of the rhombus = $\frac{\mathrm{Perimeter}}{4}$ = $\left(\frac{56}{4}\right)$ cm = 14 cm
Area of a rhombus = Base $\times$ Height

∴  Height of the rhombus = $\frac{\mathrm{Area}}{\mathrm{Base}}$= $\left(\frac{119}{14}\right)$ cm
                                          = 8.5 cm

Question 15:

The area of a rhombus is 441 cm² and its height is 17.5 cm. Find the length of each side of the rhombus.

Answer 15:

Given:
Height of the rhombus = 17.5 cm
Area of the rhombus = 441 cm²

We know:
Area of a rhombus = Base $\times$ Height

∴  Base of the rhombus =$\frac{\mathrm{Area}}{\mathrm{Height}}$ = $\left(\frac{441}{17.5}\right)$ cm = 25.2 cm
Hence, each side of a rhombus is 25.2 cm.

Question 16:

The area of a rhombus is equal to the area of a triangle whose base and the corresponding height are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the lenght of the other diagonal.

Answer 16:

Area of a triangle = $\frac{1}{2}$ $\times$ Base $\times$ Height
                          = $\left(\frac{1}{2}\times 24.8\times 16.5\right)$ cm² = 204.6 cm²
Given:
Area of the rhombus = Area of the triangle
Area of the rhombus = 204.6 cm²

Area of the rhombus = $\frac{1}{2}$ $\times$ (Product of the diagonals)
Given:
Length of one diagonal = 22 cm 
∴ Length of the other diagonal = $\left(\frac{204.6\times 2}{22}\right)$ cm
                                                  = 18.6 cm